Consider the following reaction:
2Na3PO4(aq)+3CuCl2(aq) = Cu3(PO4)2(s)+ 6NaCl(aq)
If 75mL of .175 M Na3PO4 combined with 95mL of .102M CuCl2, determine the concentration of the excess reagent.
First you identify the excess reagent. I do that the long way by taking the reactants, one at a time, to calculate the amount of product formed.
mols Na3PO4 = M x L = about 0.131
mols CuCl2 = M x L = about 0.000969,
How much NaCl could be formed with Na3PO4 and all of the CuCl2 needed. That is 0.131 mols Na3PO4 x (6 mols NaCl/2 mols Na3PO4) = 0.131 x (6/2) = approx 0.0394.
How much NaCl could be formed with CuCl2 and all of the Na3PO4 needed. That is 0.00969 x (6 mols NaCl/3 mols CuCl2) = 0.00969 x (6/3) = 0.0194
In limiting reagent problems the smaller number wins so CuCl2 is the limiting reagent.
Now follow the same procedure in determining mols Na3PO4 used by the 0.00969 mols CuCl2 and subtract from the initial amount to find the amount not used.
To determine the concentration of the excess reagent, we need to first determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's calculate the moles of Na3PO4 and CuCl2 in the given volumes.
Moles of Na3PO4 = volume (in L) × concentration in M
= 0.075 L × 0.175 M
= 0.0131 moles
Moles of CuCl2 = volume (in L) × concentration in M
= 0.095 L × 0.102 M
= 0.00969 moles
Now, let's calculate the mole ratio between Na3PO4 and CuCl2 to determine which one is the limiting reagent.
Mole ratio = Moles of Na3PO4 / Moles of CuCl2
= 0.0131 moles / 0.00969 moles
≈ 1.35
Since the mole ratio is approximately 1.35, we can conclude that Na3PO4 is the limiting reagent because it is closer to a whole number than CuCl2.
Next, we need to determine the amount of CuCl2 that reacts with the given amount of Na3PO4. From the balanced equation, we can see that the ratio of Na3PO4 to CuCl2 is 2:3. Therefore, for every 2 moles of Na3PO4, 3 moles of CuCl2 are required.
Moles of CuCl2 required = Moles of Na3PO4 × (3 moles CuCl2 / 2 moles Na3PO4)
= 0.0131 moles × (3 / 2)
= 0.0197 moles
Since the amount of CuCl2 available (0.00969 moles) is less than the amount required (0.0197 moles), CuCl2 is the limiting reagent.
Therefore, Na3PO4 is in excess.
To determine the concentration of the excess reagent (Na3PO4), we can use the following formula:
Concentration of excess reagent = (moles of excess reagent / volume of excess reagent in L)
Moles of excess Na3PO4 = Moles of Na3PO4 - Moles of Na3PO4 used
= 0.0131 moles - 0.0131 moles (since all Na3PO4 is excess)
= 0 moles
Volume of excess Na3PO4 = (volume of Na3PO4 used initially) - (volume of Na3PO4 reacted)
= 75 mL - 0 mL (since all Na3PO4 is excess)
= 75 mL = 0.075 L
Concentration of excess Na3PO4 = 0 moles / 0.075 L
= 0 M
Therefore, the concentration of the excess reagent (Na3PO4) is 0 M.
To determine the concentration of the excess reagent, we first need to identify the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed, thus determining the maximum amount of product that can be formed.
In this case, we need to compare the number of moles of Na3PO4 and CuCl2 in the given volumes and concentrations to determine the limiting reagent.
Step 1: Calculate the number of moles of Na3PO4 and CuCl2:
Number of moles of Na3PO4 = volume (in L) x concentration (in mol/L)
= 0.075 L x 0.175 mol/L
= 0.013125 mol
Number of moles of CuCl2 = volume (in L) x concentration (in mol/L)
= 0.095 L x 0.102 mol/L
= 0.00969 mol
Step 2: Determine the stoichiometric ratio of reactants:
According to the balanced equation, the ratio of Na3PO4 to CuCl2 is 2:3.
Step 3: Compare the number of moles of reactants:
Since the stoichiometric ratio is 2:3, we need to multiply the number of moles of Na3PO4 by 3 and the number of moles of CuCl2 by 2 to compare them.
Number of moles of Na3PO4 x 3 = 0.013125 mol x 3
= 0.039375 mol
Number of moles of CuCl2 x 2 = 0.00969 mol x 2
= 0.01938 mol
Step 4: Identify the limiting reagent:
The limiting reagent is the reactant with the smaller number of moles. In this case, CuCl2 has fewer moles, so it is the limiting reagent.
Step 5: Calculate the moles of the excess reagent:
To calculate the moles of the excess reagent, subtract the moles of the limiting reagent from the initial moles of that reagent.
Moles of excess Na3PO4 = Moles of Na3PO4 - Moles of Na3PO4 used in the reaction
= 0.013125 mol - 0.00969 mol (since the stoichiometric ratio is 2:3)
= 0.003435 mol
Step 6: Calculate the concentration of the excess reagent:
Concentration of excess reagent = Moles of excess reagent / Volume of excess reagent (in L)
Since the volumes of the two solutions are given as 75 mL and 95 mL respectively, we need to subtract the volumes of the limiting reagent from the total volumes to determine the volume of the excess reagent.
Volume of excess Na3PO4 = Total volume of Na3PO4 - Volume of Na3PO4 used in the reaction
= 75 mL - 0 mL (as Na3PO4 is not limiting)
Concentration of excess Na3PO4 = Moles of excess Na3PO4 / Volume of excess Na3PO4
= 0.003435 mol / (75 mL / 1000) L
= 0.0458 M
Therefore, the concentration of the excess Na3PO4 is 0.0458 M.